同格式多维数组的求和

Question

我有一个多维数组，格式如下：

array (
  0 => 
  array (
    'manual' => 1,
    'cancelled' => 1,
    'expired' => 1,
    'earned' => 1,
    'user' => 
    array (
      'user1' => 1,
      'user2' => 1,
      'user3' => 1,
      'user4' => 1,
    ),
    'transfer' => 
    array (
      'transfer1' => 
      array (
        'key1' => 1,
        'key2' => 1,
        'key3' => 1,
      ),
      'transfer2' => 
      array (
        'key5' => 1,
        'key6' => 1,
        'key7' => 1,
      ),
    ),
    'date' => '2018-03-07',
  ),
  1 => 
  array (
    'manual' => 1,
    'cancelled' => 1,
    'expired' => 1,
    'earned' => 1,
    'user' => 
    array (
      'user1' => 1,
      'user2' => 1,
      'user3' => 1,
      'user4' => 1,
    ),
    'transfer' => 
    array (
      'transfer1' => 
      array (
        'key1' => 1,
        'key2' => 1,
        'key3' => 1,
      ),
      'transfer2' => 
      array (
        'key5' => 1,
        'key6' => 1,
        'key7' => 1,
      ),
    ),
    'date' => '2018-03-08',
  ),

我需要计算具有相同索引的数组值的总和。因此，总数组应如下所示：

Array
(
    [total] => Array
        (
            [manual] => 2
            [cancelled] => 2
            [expired] => 2
            [earned] => 2
            [user] => Array
                (
                    [user1] => 2
                    [user2] => 2
                    [user3] => 2
                    [user4] => 2
                )

            [transfer] => Array
                (
                    [transfer1] => Array
                        (
                            [key1] => 2
                            [key2] => 2
                            [key3] => 2
                        )

                    [transfer2] => Array
                        (
                            [key5] => 2
                            [key6] => 2
                            [key7] => 2
                        )
        )

除了日期外，总数应该具有相同的格式，但需要显示值的总和。如何在PHP中实现这一点？我已经使用了以下代码：

$final = array_shift($input);
foreach ($final as $key => &$value){
  $value += array_sum(array_column($input, $key));
}

unset($value);

var_dump($final);

其中$input被视为第一个数组，$final是总数。我认为这仅适用于单个索引。

- Happy Coder

---

你用哪个相同的索引进行比较？ - Alive to die - Anant

你可以在这里找到解决方案：https://dev59.com/PHI_5IYBdhLWcg3wK_zE - kashalo

@kashalo 我使用了那里得到的最高票答案。但是它报错了。我认为它只能处理没有嵌套组的数组。 - Happy Coder

@AlivetoDie 实际上需要将这些索引中的值相加。不仅仅是一个，而且主数组中有许多项，不是2个。 - Happy Coder

@HappyCoder，谢谢。 - sevavietl

Answer 1

你可以使用递归函数来遍历整个“树”。如果元素是数组，则在结果数组中创建相同的数组，并为每个键运行函数。如果它是数字，只需将值添加到结果中即可。

以下是一个示例：

<?php
// Array with data
$arr=array(array('manual'=>1,'cancelled'=>1,'expired'=>1,'earned'=>1,'user'=>array('user1'=>1,'user2'=>1,'user3'=>1,'user4'=>1,),'transfer'=>array('transfer1'=>array('key1'=>1,'key2'=>1,'key3'=>1,),'transfer2'=>array('key5'=>1,'key6'=>1,'key7'=>1,)),'date'=>'2018-03-07',),array('manual'=>1,'cancelled'=>1,'expired'=>1,'earned'=>1,'user'=>array('user1'=>1,'user2'=>1,'user3'=>1,'user4'=>1,),'transfer'=>array('transfer1'=>array('key1'=>2,'key2'=>2,'key3'=>2,),'transfer2'=>array('key5'=>2,'key6'=>2,'key7'=>2,)),'date'=>'2018-03-08',)); 

//Init result array
$res=array('total'=>array());

foreach ($arr as $key=>$val) {
    //Run it for each element and store result to $res['total']
   countTotal($res['total'],$val);
}

//Show result
print_r($res);


/*
* Count totals for branch of array
* @param $res - reference to branch of result array
* @param $arr - branch of data array
*/
function countTotal(&$res,$arr) {
        foreach ($arr as $key=>$val) {
            if (is_array($val)) {
                //it's array. Create "branch" in $res and run countTotal() to calc total for it
                if (!isset($res[$key])) $res[$key]=array();
                countTotal($res[$key],$val);
            } else if (is_numeric($val)) {
                // it's number. Create "leaf" if need and add value.
                if (!isset($res[$key])) $res[$key]=0;
                $res[$key]+=$val;
            }
        }
}

你可以看到，它使用结果数组的分支引用来累加总数。

结果

    Array
    (
        [total] => Array
            (
                [manual] => 2
                [cancelled] => 2
                [expired] => 2
                [earned] => 2
                [user] => Array
                    (
                        [user1] => 2
                        [user2] => 2
                        [user3] => 2
                        [user4] => 2
                    )

                [transfer] => Array
                    (
                        [transfer1] => Array
                            (
                                [key1] => 3
                                [key2] => 3
                                [key3] => 3
                            )

                        [transfer2] => Array
                            (
                                [key5] => 3
                                [key6] => 3
                                [key7] => 3
                            )

                    )

            )

    )

在第二次“传输”中，我使用了“2”，所以总和为“3”

- ABelikov

Answer 2

你可以像下面这样编写递归函数：

function array_sum_recursive_assoc($array, &$result = [])
{
    $keys = array_keys($array[0] ?? []);

    foreach ($keys as $key) {
        foreach (($column = array_column($array, $key)) as $value) {
            if (is_array($value)) {
                $result[$key] = [];
                array_sum_recursive_assoc($column, $result[$key]);
                continue 2;
            }

            // In this case date safe guard.
            if (!is_numeric($value)) {
                continue;
            }

            $result[$key] = ($result[$key] ?? 0) + $value;
        }
    }

    return $result;
}

var_dump(array_sum_recursive_assoc($array));

这是演示。

- sevavietl